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Question 1

Topic: 9.4

(a) Red blood cells are specialised cells that transport oxygen.

State the substance in red blood cells that combines with oxygen.

(b) State the name of the component of blood that promotes blood clotting.

(c) Some students investigated the effect of immersing red blood cells in different concentrations of salt solution.

They measured the diameters of samples of red blood cells and calculated the mean.

They then immersed each red blood cell sample in a different concentration of salt solution.

After two minutes they measured and calculated the mean of the samples again.

Table 1.1 shows the results.

(i) Calculate the percentage increase in the mean diameter of red blood cells that were immersed in the 0.8% salt solution.

Give your answer to two significant figures.

Space for working.

(ii) Explain the results for the red blood cells that were immersed in the 1.8% salt solution.

(iii) State why there was no change in the mean diameter of the red blood cells immersed in the 0.9% salt solution.

(d) State why red blood cells burst when immersed in pure water but plant cells do not.

(e) State two uses of water in a plant.

▶️ Answer/Explanation
Solution

(a) haemoglobin

Explanation: Haemoglobin is the protein found in red blood cells that binds with oxygen to form oxyhaemoglobin, enabling the transport of oxygen from the lungs to tissues throughout the body.

(b) platelets

Explanation: Platelets are small cell fragments in the blood that play a crucial role in blood clotting by forming a plug at the site of injury and releasing chemicals that promote clot formation.

(c)(i) 9.3%

Working:
Initial diameter = 7.5 μm
Final diameter = 8.2 μm
Increase = 8.2 – 7.5 = 0.7 μm
Percentage increase = (0.7 / 7.5) × 100 = 9.333…% ≈ 9.3% (to 2 significant figures)

(c)(ii)

Explanation: The 1.8% salt solution has a lower water potential than the red blood cells. This causes water to move out of the cells by osmosis, across the partially permeable membrane, from an area of higher water potential (inside the cell) to lower water potential (the salt solution). As a result, the cells shrink, decreasing their mean diameter from 7.5 μm to 6.0 μm.

(c)(iii) The water potential of the red blood cells is the same as the 0.9% salt solution.

Explanation: When the water potential inside and outside the cell is equal, there is no net movement of water across the cell membrane. This isotonic condition means the cells neither gain nor lose water, so their diameter remains unchanged.

(d) Red blood cells do not have a rigid cell wall.

Explanation: In pure water, red blood cells take in water by osmosis and burst (lyse) because they lack a cell wall to prevent over-expansion. Plant cells have a rigid cell wall that resists the inward pressure of water, preventing bursting and instead making the cell turgid.

(e) Any two from:

  • Transport of ions/sucrose (translocation)
  • As a reactant in photosynthesis
  • As a solvent for metabolic reactions
  • Maintaining turgidity of cells (support)
  • Medium for enzyme reactions

Explanation: Water is essential in plants for various functions. It acts as a transport medium in the xylem and phloem, is a raw material in photosynthesis, provides structural support through turgor pressure, and serves as a solvent for biochemical reactions.

Question 2

Topic: 18.2

Fig. 2.1 is a photograph of some leaves of a water lily, which is a hydrophyte. The water lily has adaptive features that are found in many different hydrophytes.

(a) Describe what is meant by an adaptive feature.

(b) Fig. 2.2 is a photomicrograph of a cross-section of a part of a water lily leaf.

(i) State the names of the parts labelled A, B and C in Fig. 2.2.

(ii) Explain how part C in Fig. 2.2 adapts the hydrophyte for its environment.

(c) A scientist calculated the mean number of stomata per mm² in the upper and lower epidermis in tomato plants and water lily plants. Tomato plants are a type of terrestrial plant.

Table 2.1 shows the results.

(i) Compare and explain the differences in the mean number of stomata in a tomato plant and in a water lily plant.

(ii) State the name of the cells that control the opening and closing of stomata.

▶️ Answer/Explanation
Solution

(a) An inherited (structural) feature that helps an organism to survive or reproduce in its environment.

Explanation: Adaptive features are characteristics that organisms develop over generations through natural selection. These features enhance the organism’s ability to survive in its specific environment, such as the water lily’s floating leaves that allow it to thrive in aquatic conditions.

(b)(i)

A – palisade mesophyll

B – spongy mesophyll

C – air spaces

Explanation: The palisade mesophyll (A) contains tightly packed chloroplast-rich cells for photosynthesis. The spongy mesophyll (B) has loosely arranged cells with air spaces (C) between them to facilitate gas exchange.

(b)(ii) The air spaces reduce the density of the leaf, allowing it to float on water and be near the surface where it can access light and carbon dioxide for photosynthesis.

Explanation: The extensive air spaces in water lily leaves serve multiple adaptive functions. They make the leaf buoyant, keeping it at the water’s surface where sunlight is abundant. This positioning also ensures access to atmospheric CO₂ for photosynthesis while the lower surface remains submerged.

(c)(i)

Tomato plants have more stomata on their lower epidermis (129/mm²) than upper epidermis (10/mm²), while water lilies have all stomata on the upper epidermis (475/mm²) and none on the lower epidermis.

Tomatoes reduce water loss by having most stomata on the shaded lower surface. Water lilies, floating on water, don’t need to conserve water and have stomata only on the upper surface exposed to air.

Explanation: Terrestrial plants like tomatoes evolved stomatal distribution to minimize water loss through transpiration, hence more stomata on the cooler, shaded lower surface. Hydrophytes like water lilies, with constant water supply, maximize gas exchange through numerous upper-surface stomata since their lower surface is underwater where stomata would be useless.

(c)(ii) Guard cells

Explanation: Guard cells are specialized kidney-shaped cells that flank each stoma. By changing their turgor pressure through osmotic water movement, they can open or close the stomatal pore, regulating gas exchange and transpiration.

Question 3

Topic: 14.1

Fig. 3.1 is a diagram of the junction between two neurones in a healthy person.

Fig. 3.2 is a diagram of the junction between the same two neurones in a person who has Parkinson’s disease. This disease affects the nervous system.

 

(a) Identify the parts labelled X and Y in Fig. 3.2.

(b) Parkinson’s disease affects neurones in the brain that are responsible for movement.

Using the information in Fig. 3.1 and Fig. 3.2, suggest and explain the effect of Parkinson’s disease on a person’s movement.

(c) Describe two ways nervous control differs from hormonal control.

(d) (i) The shape of the receptor proteins shown in Fig. 3.1 and Fig. 3.2 is important for their function.

Explain how the shape of the receptor proteins is determined.

(ii) Cell membranes also contain protein carriers. Describe the role of protein carriers.

▶️ Answer/Explanation
Solution

(a) X – vesicle (membrane); Y – synapse/synaptic gap

Explanation: In Fig. 3.2, X represents the membrane-bound vesicles that contain neurotransmitters, which are crucial for signal transmission between neurones. Y is the synaptic gap, the physical space between two neurones where neurotransmitters diffuse to carry the signal from one neurone to the next.

(b)

Explanation: Parkinson’s disease significantly impairs movement due to several neurological changes:

  1. Movement becomes slower and less coordinated because there are fewer vesicles containing neurotransmitters.
  2. The reduced number of neurotransmitters means fewer signals are transmitted across the synaptic gap.
  3. With fewer neurotransmitters binding to receptor proteins, the postsynaptic neurone receives weaker signals.
  4. This results in slower reflexes and difficulty initiating or controlling movements.
  5. The overall effect is impaired motor function, characteristic of Parkinson’s symptoms like tremors and rigidity.

(c)

  1. Nervous control is much faster than hormonal control, with responses occurring in milliseconds compared to seconds or minutes for hormones.
  2. Nervous control is localized to specific areas through neural pathways, while hormonal control has widespread effects throughout the body via the bloodstream.

Explanation: The nervous system uses electrical impulses that travel quickly along neurones, allowing for rapid responses. In contrast, hormones are chemical messengers that travel through the blood, resulting in slower but more sustained effects. Additionally, nervous control targets specific muscles or glands, whereas hormones can affect multiple target organs simultaneously.

(d)(i)

Explanation: The shape of receptor proteins is determined by:

  1. The specific sequence of bases in the gene/DNA that codes for the protein.
  2. This DNA sequence is transcribed into mRNA and translated into a specific sequence of amino acids.
  3. The amino acid sequence then folds into a unique three-dimensional shape that determines the protein’s function.
  4. This precise shape allows the receptor to bind specifically to certain neurotransmitters.

(d)(ii)

Explanation: Protein carriers in cell membranes have crucial roles:

  1. They actively transport specific molecules or ions across the membrane against concentration gradients.
  2. This process requires energy from ATP produced during respiration.
  3. They enable the movement of substances that cannot diffuse freely through the membrane.
  4. Examples include transporting glucose into cells or pumping sodium and potassium ions in nerve impulses.
Question 4

Topic: 6.1

A scientist investigated the effect of temperature on the rate of photosynthesis in one species of plant. Photosynthesis involves enzyme-controlled reactions. Discs were cut from a leaf and kept at different temperatures. The total surface area of the discs was kept the same for each temperature. The volume of oxygen that was produced by the leaf discs was measured and used to estimate the rate of photosynthesis.

The results are shown in Fig. 4.1.

 

(a) Identify the optimum temperature for photosynthesis in Fig. 4.1.

(b) Explain the results shown in Fig. 4.1.

(c) Carbon dioxide was supplied in excess at each temperature during the investigation. Explain why.

(d) Suggest why not all of the oxygen produced by the leaf is released.

(e) Describe the role of chlorophyll in photosynthesis.

▶️ Answer/Explanation
Solution

(a) 40 °C

Explanation: The optimum temperature is where the rate of photosynthesis is highest. From the graph, this occurs at 40 °C, as it is the peak point before the rate begins to decline.

(b)

Explanation: The graph shows an initial increase in the rate of photosynthesis as temperature rises, peaking at 40 °C, followed by a sharp decline. Here’s why:

  • Before 40 °C: Increasing temperature boosts the kinetic energy of molecules, leading to more frequent collisions between enzymes and substrates. This forms more enzyme-substrate complexes, increasing the rate of photosynthesis. Temperature is the limiting factor here.
  • At 40 °C: This is the optimum temperature where enzyme activity is highest, resulting in the maximum rate of photosynthesis.
  • After 40 °C: Higher temperatures cause enzymes to denature. The active site changes shape, substrates no longer fit, and fewer enzyme-substrate complexes form. This reduces the rate of photosynthesis, eventually stopping it entirely.

(c)

Explanation: Carbon dioxide was supplied in excess to ensure it was not a limiting factor. By controlling this variable, the scientist could isolate the effect of temperature on the rate of photosynthesis. This way, any changes observed could be confidently attributed to temperature variations alone.

(d)

Explanation: Not all oxygen produced is released because some of it is used by the plant for aerobic respiration. Additionally, a portion may remain trapped in the air spaces within the leaf tissues.

(e)

Explanation: Chlorophyll plays two critical roles in photosynthesis:

  • It absorbs light energy, primarily from the red and blue regions of the visible spectrum, which is essential for driving the photosynthetic process.
  • It converts light energy into chemical energy, which is used to synthesize glucose from carbon dioxide and water. This conversion is the foundation of the food chain, as glucose serves as an energy source for the plant and other organisms.
Question 5

Topic: 9.1

(a) Describe two ways in which the circulatory system of a fish is different from the circulatory system of a mammal.

(b) Explain the advantages of a double circulatory system.

(c) Fig. 5.1 shows part of the circulatory system of a mammal.

 

(i) State the letter from Fig. 5.1 that identifies where these processes occur:

absorption of the products of digestion into the blood 
excretion of carbon dioxide from the body 
formation of urine 
production of bile.

(ii) Identify the name of the blood vessel labelled X in Fig. 5.1.

(d) Describe the role of the liver in excretion.

▶️ Answer/Explanation
Solution

(a)

1. Fish have single circulation (blood flows through the heart once per body circuit) while mammals have double circulation (blood flows through the heart twice per body circuit).

2. Fish hearts have two chambers (one atrium and one ventricle) while mammalian hearts have four chambers (two atria and two ventricles).

3. Fish hearts have no septum (allowing mixing of oxygenated and deoxygenated blood) while mammalian hearts have a complete septum separating oxygenated and deoxygenated blood.

Explanation: The fish circulatory system is simpler and less efficient than mammals’. The single circulation means blood pressure drops after passing through gills, while mammals maintain high pressure to body tissues due to double circulation. The four-chambered heart in mammals completely separates oxygenated and deoxygenated blood, making oxygen delivery more efficient.

(b)

1. Prevents mixing of oxygenated and deoxygenated blood, maintaining high oxygen delivery efficiency.

2. Allows blood to flow at high pressure to body tissues, enabling rapid nutrient and oxygen delivery.

3. Enables efficient waste removal (carbon dioxide, urea) from tissues.

4. Maintains lower pressure in pulmonary circuit to protect delicate lung capillaries.

5. Supports higher metabolic rates needed for endothermy (warm-bloodedness) in mammals.

Explanation: The double circulatory system separates systemic and pulmonary circuits, allowing different pressure optimizations. The high pressure systemic circuit delivers oxygen effectively to all tissues, while the lower pressure pulmonary circuit prevents lung damage. This separation is crucial for maintaining the high metabolic demands of mammals.

(c)(i)

Absorption of digestion products: C (small intestine)
Excretion of CO₂: A (lungs)
Formation of urine: D (kidneys)
Production of bile: B (liver)

Explanation: The small intestine (C) absorbs digested nutrients into blood capillaries. Lungs (A) excrete carbon dioxide during gas exchange. Kidneys (D) filter blood to form urine. The liver (B) produces bile which aids fat digestion.

(c)(ii) Hepatic portal vein

Explanation: The hepatic portal vein carries nutrient-rich blood from the digestive organs to the liver for processing before this blood enters general circulation.

(d)

1. Deaminates excess amino acids, removing toxic amino groups which are converted to urea.

2. Converts ammonia (toxic) into urea (less toxic) which is excreted by kidneys.

3. Breaks down old red blood cells, processing hemoglobin into bile pigments.

4. Detoxifies harmful substances like alcohol and drugs by chemical modification.

5. Processes lactic acid produced during anaerobic respiration.

Explanation: The liver acts as the body’s main chemical processing plant for excretion. By converting toxic substances like ammonia into less harmful ones like urea, and breaking down various waste products, it plays a crucial role in maintaining the body’s internal environment. The urea produced is then filtered out by the kidneys.

Question 6

Topic: 16.5

(a) State the names of two hormones released by the ovaries.

(b) During pregnancy, antibodies are acquired by the fetus from the mother.

State the organ the antibodies cross to reach the fetus.

(c) A baby was breastfed for six months. The concentration of antibodies in the baby’s blood obtained from breast milk and the concentration of antibodies made by the baby itself were measured.

Fig. 6.1 shows the results.

 

(i) Complete the sentences to describe the changes in antibody concentration in the baby.

After birth the concentration of antibodies acquired from the mother decreases rapidly to 0 arbitrary units at ………………………………… months.

Antibodies start being produced by cells called ………………………………… in the baby immediately after birth.

The total concentration of antibodies in the baby from both sources is ………………………………… arbitrary units at 4 months.

The concentration of antibodies acquired from the mother and the concentration of antibodies produced by the baby are the same at ………………………………… months.

(ii) Describe the benefits of breastfeeding a baby for the first six months of life.

(iii) State two ways, other than breastfeeding, that a baby can acquire immunity.

(d) Explain the importance of the shape of an antibody.

(e) Antibodies are proteins.

State the chemical elements present in all proteins that are also found in carbohydrates and fats.

▶️ Answer/Explanation
Solution

(a) oestrogen; progesterone

Explanation: The ovaries release two main hormones – oestrogen and progesterone. Oestrogen is responsible for the development of female secondary sexual characteristics and regulation of the menstrual cycle, while progesterone prepares the uterus for pregnancy and maintains the uterine lining.

(b) placenta

Explanation: The placenta is the specialized organ that forms during pregnancy to allow the transfer of antibodies and nutrients from the mother’s bloodstream to the fetus while keeping their blood supplies separate.

(c)(i) 6; lymphocytes; 56; 3

Explanation:

  • The graph shows maternal antibodies decrease to zero by 6 months.
  • Lymphocytes are the white blood cells that produce antibodies.
  • At 4 months, the total antibodies (maternal + baby’s own) add up to 56 arbitrary units.
  • The two lines intersect at 3 months when concentrations are equal.

(c)(ii) Any two from:

  • Provides passive immunity by supplying maternal antibodies
  • Contains all necessary nutrients for baby’s growth
  • Helps protect against infections and diseases
  • Creates bonding between mother and baby
  • Is always at the right temperature and sterile

(c)(iii) Any two from:

  • Antibodies received across placenta during pregnancy
  • Vaccination (active immunity)
  • Direct antibody injections (passive immunity)
  • Natural infection leading to antibody production

(d) Key points:

  • Each pathogen has unique surface antigens
  • Antibodies have complementary shapes to specific antigens
  • This allows precise binding to mark pathogens for destruction
  • The specific shape enables the immune system to recognize and neutralize specific threats

(e) carbon, hydrogen, oxygen

Explanation: These three elements are common to all proteins, carbohydrates and fats. Proteins additionally contain nitrogen and sometimes sulfur, while carbohydrates and fats contain only C, H and O.

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